def gemstones(arr):
    gems = set(arr[0])
    
    for stone in arr[1:]:
        gems = set(stone).intersection(gems)
    
    return len(gems)

int gemstones(vector<string> arr) {
    int result = 0;
    string liste = "abcdefghijklmnopqrstuvwxyz";
    for(int j = 0; j < liste.size(); j++){
        int cp = 0;
        for(int i = 0; i < arr.size(); i++){
            if(arr[i].find(liste[j]) != string::npos) cp++;
        }
        if(cp == arr.size())result++;
    }
    return result;
}

string getShortest(vector<string> arr) {
    string result = arr[0];
    for(int i = 1; i < arr.size(); i++){
        if(arr[i].size() < result.size()) result = arr[i];
    }
    return result;
}
int gemstones(vector<string> arr) {
    int result = 0;
    string shortest = getShortest(arr);
    set<char> liste(shortest.begin(), shortest.end());
    for(auto j = liste.begin(); j != liste.end(); j++){
        int cp = 0;
        for(int i = 0; i < arr.size(); i++){
            if(arr[i].find(*j) != string::npos) cp++;
        }
        if(cp == arr.size())result++;
    }
    return result;
}

public static int gemstones(List<String> arr) {
        //given an array of strings, determine the number of characters that appear in all strings 
        //get the set of chars in the first string
        Set<Character> commonSet = new HashSet<>();
        for(char mineral: arr.get(0).toCharArray()){
            commonSet.add(mineral);
        }
        //iterate through every other string, add their chars to currentset, and use retain all to find the intersection
        for(int i = 1; i < arr.size(); i++){
            Set<Character> currentSet = new HashSet<>();
            for(char mineral: arr.get(i).toCharArray()){
                currentSet.add(mineral);
            }
            commonSet.retainAll(currentSet); //intersection to find common chars
        }
        //return size of the og set
        return commonSet.size();
    }

def gemstones(arr):
    m=set(arr[0])
    for i in (arr):
        m=set(i).intersection(m)
    return len(m)